Constant expressions are not guaranteed to be evaluated at compile time, we only have a non-normative quote from draft C++ standard section 5.19 Constant expressions that says this though:
[…]>[ Note: Constant expressions can be evaluated during
translation.—end note ]
You can assign the result to constexpr variable to be sure it is evaluated at compile time, we can see this from Bjarne Stroustrup’s C++11 reference which says (emphasis mine):
In addition to be able to evaluate expressions at compile time, we
want to be able to require expressions to be evaluated at compile
time; constexpr in front of a variable definition does that (and
implies const):
For example:
constexpr int len1 = length("abcd") ;
Bjarne Stroustrup gives a summary of when we can assure compile time evaluation in this isocpp blog entry and says:
[…]The correct answer – as stated
by Herb – is that according to the standard a constexpr function may
be evaluated at compiler time or run time unless it is used as a
constant expression, in which case it must be evaluated at
compile-time. To guarantee compile-time evaluation, we must either use
it where a constant expression is required (e.g., as an array bound or
as a case label) or use it to initialize a constexpr. I would hope
that no self-respecting compiler would miss the optimization
opportunity to do what I originally said: “A constexpr function is
evaluated at compile time if all its arguments are constant
expressions.”
So this outlines two cases where it should be evaluated at compile time:
- Use it where a constant expression is required, this would seem to be anywhere in the draft standard where the phrase
shall be ... converted constant expressionorshall be ... constant expressionis used, such as an array bound. - Use it to initialize a
constexpras I outline above.